Product Rule: $X \equiv x \text{ mod } m$ and $Y \equiv y \text{ mod } m \implies X \cdot Y \equiv x \cdot y \text{ mod } m$.
Proof: See here. Another way is to simply expand out the terms, do the calculations using normal arithmetic, and then express the result back in modular arithmetic.
Expansion:
$\begin{align}X &\equiv x \text{ mod } m \iff X = k_1 \cdot m + x \text{ where } k_1 \in \mathbb{Z} \\ Y &\equiv y \text{ mod } m \iff Y = k_2 \cdot m + y \text{ where } k_2 \in \mathbb{Z} \end{align}$
Calculation using normal arithmetic:
$\begin{align}X \cdot Y &= (k_1 \cdot m + x) \cdot (k_2 \cdot m + y)\\ &=\underbrace{(k_1k_2m + k_1y + k_2x)}_{k \in \mathbb{Z}}\cdot m + xy \\ &= k \cdot m + xy\end{align}$
Result:
$X \cdot Y = k \cdot m + xy \iff X \cdot Y \equiv xy \text{ mod } m$
Conceptually, multiplication between two integers $X, Y$ can be thought of as repeated addition (assuming $Y > 0$), so it's not surprising that a property of integer addition (congruence sum rule) applied repeatedly may lead to a property of integer multiplication.
mod m:
$\begin{align}X \cdot Y &\equiv \underbrace{X + X + X + \cdots + X}_\text{Y times}\\& \equiv \underbrace{x + x + x + \cdots + x}_\text{Y times} &\text{sum rule applied Y-1 times}\\ &\equiv x \cdot Y = Y \cdot x \\&\equiv \underbrace{Y + Y + Y + \cdots + Y}_\text{x times}\\& \equiv \underbrace{y + y + y + \cdots + y}_\text{x times} &\text{sum rule applied x-1 times}\\ &\equiv yx = xy\end{align}$